NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question3(viii)
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 3 (viii)
Question
In an AP: given an = 4, d = 2, Sn = –14, find n and a.
Given
an = 4, d = 2, Sn = –14
To Find
n and a
Formula
The nth term of arithmetic progression is given by an = a + (n-1)d
Sum of an arithmetic progression = Sn = n2(2a + (n-1)d)
Solution
Using an = a + (n-1)d
⇒ 4 = a + (n-1)2
⇒ 4 = a + 2n - 2
⇒ a = 6 - 2n
Now, using the formula Sn = n2(2a + (n - 1)d), we have:
⇒ -14 = n2(2a + (n-1)2)
⇒ -28 = n(2a + 2n - 2)
Substituting a = 6 - 2n
⇒ -28 = n(2(6 - 2n) + 2n - 2)
⇒ -28 = n(12 - 4n + 2n - 2)
⇒ -28 = n(10 - 2n)
⇒ -28 = 10n - 2n2
⇒ 2n2 - 10n - 28 = 0
⇒ n2 - 5n - 14 = 0
Solving the quadratic equation: (n-7)(n+2) = 0
⇒ n = 7 or n = -2.
Since n cannot be negative, n = 7
Substituting n = 7 in a = 6 - 2n
⇒ a = 6 - 2(7)
= 6 - 14
= -8
Result
n = 7 and a = -8
Next question solution
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 3 (ix)
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